Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    double[] p = new double[N];
    for (int i = 0; i < N; i++)
      p[i] = sc.nextDouble();
    sc.close();

    double[][] dp = new double[N][N + 1];
    dp[0][0] = 1d - p[0];
    dp[0][1] = p[0];
    for (int i = 0; i < N - 1; i++) {
      for (int j = 0; j < i + 2; j++) {
        dp[i + 1][j] += dp[i][j] * (1d - p[i + 1]);
        dp[i + 1][j + 1] += dp[i][j] * p[i + 1];
      }
    }

    double ans = 0;
    for (int i = (N + 1) / 2; i < N + 1; i++) ans += dp[N - 1][i];
    System.out.println(ans);
  }
}